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The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.5 ✕ 10−2 s-1. (a) if we begin with an initial so2cl2 pressure of 430. torr, what is the pressure of this substance after 70. s?

Respuesta :

W0lf93
It is a first order process. So the partial pressure at one point is related to partial pressure at other point in time. So the gas-phase reaction is  
Ln PSO2Cl2 = Ln (Initial PSO2Cl2) - k x t 
Initial Pressure P = 430 = Initial Ln PSO2Cl2. 
Time t = 65 s 
Constant k = 4.5 âś• 10â’2 s-1 
Ln PSO2Cl2 = Ln 430 - (4.5 âś• 10^â’2 x 70) 
Ln PSO2Cl2 = 6.063 - 3.15 
Ln PSO2Cl2 = 2.913 
PSO2Cl2 = 18.43 torr 
The pressure of substance after 70 s = 18.43 torr

Answer:

18.43 Torr is the pressure of this substance after 70seconds.

Explanation:

[tex]SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)[/tex]

Rate of the reaction ,k= [tex]4.5\times 10^{-2} s^{-1}[/tex]

Integrated rate equation for first order kinetics in gas phase is given as:

[tex]k=\frac{2.303}{t}\log\frac{p_o}{p}[/tex]

p= pressure of the gas at given time t.

[tex]p_o[/tex] = Initial pressure of the gas

When, t = 70 sec

[tex]p_o=430 torr[/tex]

[tex]4.5\times 10^{-2} s^{-1}=\frac{2.303}{70 s}\log\frac{p_o}{p}[/tex]

p = 18.43 Torr

18.43 Torr is the pressure of this substance after 70 seconds.

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